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Lecture OutlineIntroductionAC Circuits Series LCReR CircuiteC CircuitPowerPoint PresentationeL CircuitSlide 9Lecture 20, ACT 1Slide 11PhasorsPhasors for L,C,RLecture 20, ACT 2Slide 15Series LCR AC CircuitPhasors: LCRSlide 18Lecture 20, ACT 3Slide 20Phasors:LCRPhasors:TipsSlide 23Physics 1304: Lecture 17, Pg 1 LCRimRimL imCmPhysics 1304: Lecture 17, Pg 2Lecture OutlineLecture OutlineIntroduction:•Can a circuit be built with real components (ie resistance) which sustains the oscillations we saw in an LC circuit?•Phases for driven circuits with R, C, and LPhasors:•Defined as a rotating vector•Phase difference between current & voltage for Resistors, Capacitors, and Inductors. Application to Driven Series LCR Circuit:•General solutionText Reference: Chapter 33.1-4Physics 1304: Lecture 17, Pg 3IntroductionIntroduction•However, any real attempt to construct a LC circuit must account for the resistance of the inductor. This resistance will cause oscillations to damp out.•Question: Is there any way to modify the circuit above to sustain the oscillations without damping?Answer: Yes, if we can supply energy at the rate the resistor dissipates it! How? A sinusoidally varying emf (AC generator) will sustain sinusoidal current oscillations!•Last time we discovered that a LC circuit was a natural oscillator.LC+ +--RPhysics 1304: Lecture 17, Pg 4AC CircuitsAC CircuitsSeries LCRSeries LCR•Statement of problem:Given  = msint , find i(t).Everything else will follow.•Procedure: start with loop equation?•We could solve this equation in the same manner we did for the LCR damped circuit. Rather than slog through the algebra, we will take a different approach which uses phasors.LCRLd QdtQCRdQdttm22   sinPhysics 1304: Lecture 17, Pg 5R CircuitR CircuitBefore introducing phasors, per se, begin by considering simple circuits with one element (R,C, or L) in addition to the driving emf.•Begin with R: Loop eqn gives:Voltage across R in phase with current through RiRRV Ri tR R m  siniRtRmsin00VRtm m00tiRmRmRNote: this is always, always, true… always.Physics 1304: Lecture 17, Pg 6C CircuitC Circuit•Now consider C: Loop eqn gives:CiCVoltage across C lags current through C by one-quarter cycle (90).VQCtC m  sinQ C tm  sinidQdtC tC m   cos00tVCm mt0iC0 Cm  CmIs this always true? YESPhysics 1304: Lecture 17, Pg 7)2/sin()cos( tt)()( btftf )(tft)( btf bPhysics 1304: Lecture 17, Pg 8L CircuitL Circuit•Now consider L: Loop eqn gives:Voltage across L leads current through L by one-quarter cycle (90).LiLV LdidttLLm  sindiLtdtLmsini diLtL LmcostVL00m mt0iL0mLmLYes, yes, but how to remember? 2/tsinLmPhysics 1304: Lecture 17, Pg 9VL leads ILVC lags ICHi kids, I’m Eli and I’ll help you learn physics !…we’ll see about that1Introducing...Physics 1304: Lecture 17, Pg 10Lecture 20, ACT 1Lecture 20, ACT 1A circuit consisting of capacitor C and voltage source  is constructed as shown. The graph shows the voltage presented to the capacitor as a function of time. Which of the following graphs best represents the time dependence of the current i in the circuit? tC(a)(b)(c)itittiPhysics 1304: Lecture 17, Pg 11Lecture 20, ACT 1Lecture 20, ACT 1A circuit consisting of capacitor C and voltage source  is constructed as shown. The graph shows the voltage presented to the capacitor as a function of time. Which of the following graphs best represents the time dependence of the current i in the circuit? tC(a)(b)(c)ititti• This curve says the current is in phase with the applied voltage. • This is only possible when the load is purely resistive.• Nope!• This curve says the current is the mirror image of the applied voltage. • This is not right, but it correctly recognizes that when  is constant, Q is not changing!• This is the correct curve.• The  time dependence is the time dependence of the charge on the capacitor. ie  = VC = Q/C• The current i = dQ/dt.• This curve is the derivative of the applied voltage curve.Physics 1304: Lecture 17, Pg 12PhasorsPhasorsA phasor is a vector whose magnitude is the maximum value of a quantity (eg V or I) and which rotates counterclockwise in a 2-d plane with angular velocity . Recall uniform circular motion:The projections of r (on the vertical y axis) execute sinusoidal oscillation. iLtLmcosi C tC m  cosiRtRmsinx r t cosy r t sinV Ri tR R m  s in•R: V in phase with iVQCtC m  sin•C: V lags i by 90V LdidttLLm  sin•L: V leads i by 90xyyPhysics 1304: Lecture 17, Pg 13Phasors for L,C,RPhasors for L,C,RitVRitVLitVCSuppose:i i tm sinV Ri tR m sinVCi tC m1cosV Li tL m costiVR0VC0ii0VL2Physics 1304: Lecture 17, Pg 14Lecture 20, ACT 2Lecture 20, ACT 2A series LCR circuit driven by emf  = 0sint produces a current i=imsin(t-). The phasor diagram for the current at t=0 is shown to the right.At which of the following times is VC, the magnitude of the voltage across the capacitor, a maximum? LCRit=0(a)(b)(c)it=0it=tbit=tcPhysics 1304: Lecture 17, Pg 15Lecture 20, ACT 2Lecture 20, ACT 2A series LCR circuit driven by emf  = 0sint produces a current i=imsin(t-). The phasor diagram for the current at t=0 is shown to the right.At which of the following times is VC, the magnitude of the voltage across the capacitor, a maximum? LCRit=0• The phasor diagram for the driven series RLC circuit always has the voltage across the capacitor lagging the current by 90. The white arrow therefore represents the phasor for VC.(a)(b)(c)it=0it=tbit=tcVCVCVC• The voltage across the capacitor is given by the projection of the VC phasor along the vertical axis.• The maximum vertical projection occurs when the current is zero (tb).Physics 1304: Lecture 17, Pg 16Series LCRSeries LCRAC CircuitAC Circuit•Back to the original problem: the loop equation gives:Here all unknowns, (im,) , must be found from the loop eqn; the initial


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SMU PHYS 1304 - Lecture Notes

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